Let $d\ge 1$ be an integer and let $\vec{A}:=\left\{ A_i \right\}_{i=1}^d$ be real numbers. We consider a following integral:
\begin{equation}
{\mathfrak I}^{(d)}(\vec{A}):=\int\limits_0^\infty e^{-u^2}\left[ \prod_{i=1}^d \operatorname{erf}(A_i u) \right] du
\end{equation}
By expanding the error functions in Taylor series and then integrating term by term we found the answer for $d=1$ and $d=2$. We have:
\begin{eqnarray}
\sqrt{\pi} {\mathfrak I}^{(d)}(\vec{A}) =
\begin{cases}
\arctan(A_1) & \text{if $d=1$}\\[4pt]
\arctan\left(\frac{A_1 A_2}{\sqrt{1+A_1^2+A_2^2}}\right) & \text{if $d=2$}
\end{cases}
\end{eqnarray}
Now the question is how do we derive the result for arbitrary values of $d$?
Friday, 26 December 2014
integration - An integral involving error functions and a Gaussian
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